Question

If the solution curve of the differential equation $((tan−1y)−x)dy=(1+y^2)dx$ passes through the point $(1, 0)$, then the abscissa of the point on the curve whose ordinate is $tan(1)$, is

• A. $2e$
• B. $\frac{2}{e}$
• C. $e$
• D. $\frac{1}{e}$

Answer 1

Answer: (B)

$\frac{2}{e}$

Answer 2

$((tan−1y)−x)dy=(1+y^2)dx$

$\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{tan^{−1}y}{1+y^2}$

$I.F.=e^{∫\frac{1}{1+y^2}dy}=e^{tan^{−1}y}$

∴ Solution

$x.e^{tan^{-1}y}∫\frac{e^{tan^{-1}}y^{tan^{-1}}y}{1+y^2}dy$

Let

$e^{tan^{−1}}y=t$

$\frac{e^{tan^{−1}y}}{1+y^2}=dt$

=$xe^{tan^{−1}y}∫ ln\; tdt =t \;ln t–t+c$

=$xe^{tan^{−1}y}=e^{tan^{−1}y}tan^{−1}y−e^{tan^{-1}y}+c…(i)$

∵ It passes through $(1, 0) ⇒ c = 2$

Now put $y = tan1$, then

$ex = e – e \+ 2$

$⇒x=\frac{2}{e}$